# Quiz: Civil Engineering 17 May 2021

Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/3 mark
Time: 20 Minutes

Q1. A clay layer of thickness 10 cm and initial void ratio 0.5 undergoes settlement that final void ratio is 0.2, the settlement of the layer in cm is:
(a) 1
(b) 1.5
(c) 2
(d) 2.5

Q2. A soil sample has hulk density of 21 kN/m³ and water content 8% Dry density of the sample is –
(a) 19.00 kN/m²
(b) 18.44 kN/m²
(c) 22.68 kN/m²
(d) 19.44 kN/m²

Q3. In an experiment it is found that the bulk modulus of a material is equal to its shear modulus. The poison’s ratio is
(a) 0.150
(b) 0.125
(c) 0.250
(d) 0.375

Q4. In a BOD test, 5 ml of waste is added to 295 ml of aerated pure water. Initial dissolved oxygen (DO) content of the diluted sample is 7.8 mg/l. after 5 days of incubation at 20°C, the DO content of the sample is reduced to 4.4 mg/l. the BOD of the waste water is
(a) 196 mg/l
(b) 200 mg/l
(c) 204 mg/l
(d) 208 mg/l

Q5. Water flows at a rate of 10 m³/s in a rectangular channel 3 m wide. The critical depth of flow is
(a) 1.13 m
(b) 2 m
(c) 1.45 m
(d) 1.04 m

Q6. The circulation ‘Γ’ around a circle of radius 2 units for the velocity field u = 2x+3y and v= -2y is
(a) -6π units
(b) -12π units
(c) -18π units
(d) -24π units

Q7. An isolated T-beam is used as a walkway. The beam is simply supported with an effective span of 6 m. the effective width of flange, for the cross-section shown in figure is

(a) 900 mm
(b) 1000 mm
(c) 1259 mm
(d) 2200 mm

Q8. An observer standing on a signal post between two forts situated in a straight line sees the top of both the forts. If the signal post acts as point of horizon. What is the distance between two forts. If fort A and B are 3 km and 5 km above the sea level respectively?
(a) 280 km
(b) 480 km
(c) 380 km
(d) 580 km

Q9. If the working stress in bearing in power driven rivets is 300 N\/mm^2 and a double riveted double cover butt joint is used to connect plates of 12 mm thick, then the strength of rivet in bearing is (Given; The nominal diameter of rivet= 22 mm)
(a) 84.6 kN
(b) 54.6 kN
(c) 72.6 kN
(d) 68.4 Kn

Q10. The principal strains at a point are + 800 × 10^(-6) cm/cm, + 400 × 10^(-6) cm/cm and -1200 × 10^(-6) cm/cm. The volumetric strain is equal to
(a) +1200×10^(-6) cm\/cm
(b) + 800 × 10^(-6) cm/cm
(c) -1200×10^(-6) cm\/cm
(d) zero
Solutions
S1. Ans.(c)
Sol. Thickness of clay later (H) = 10 cm.
Initial void ratio (e_0) = 0.5
Final void ratio (e_f)= 0.2
Change in void ratio ∆e = e_0-e_f
=0.5 – 0.2
=0.3
Settlement of clay layer (∆H)= ?
We know,
∆H/H=∆e/(1+e_0 )
∆H=(0.3×10)/(1+0.5 )
=30/15
▭(∆H=2cm)

S2. Ans.(d)
Sol. Given,
γ_b=21 kN\/m^2
w=8%=0.08
γ_d= ?
We know,
γ_d=γ_b/(1+w)
=21/(1+0.08)
▭(γ_d=19.44 kN\/m^3 )

S3. Ans.(b)
Sol. Given, Bulk modulus (K) = Shear modulus (G)
We know,
E=2G(1+μ)
G=E/2(1+μ) —–(1)
E=3K (1-2μ)
K=E/3(1-2μ) —-(2)
From Eq^n (1) & (2)
E/3(1-2μ) =E/2(1+μ)
2+2μ=3-6μ
8μ=1
▭(μ=1/8=0.125)

S4. Ans.(c)
Sol.
(DO_i )=7.8 mg\/l
(DO_f )=4.4 mg\/l
(BOD)_5= ?
(BOD)_5=(DO_i-DO_f )×Dilution factor
=(7.8-4.4)×300/5
▭((BOD)_5=204 mg\/l )

S5. Ans.(d)
Sol.
Q=10m^3 \/sec.
B=3 m
y_c= ?
q=Q/B=10/3=3.33 m^3 \/sec.
Now critical depth (y_c )
y_c=(q^2/g)^(1\/3)
=[(3.33)^2/9.8]^(1\/3)
▭(y_c=1.04m)

S6. Ans.(b)
u = 2x+3y
V= -2y
Circulation (Γ)= ?
Circulation (Γ)= Vorticity×Area
=(∂V/∂x-∂u/∂y)×Area
=(0-3)×π×(2)^2
= -12π units.

S7. Ans.(a)
Sol. effective span (l_o )=6m=6000 mm.
Width (B) = 1000 mm.
bw = 300 mm.
Effective width of flange (b_f )= ?
For isolated T-beam
b_f=l_0/(l_0/B+4 )+bw
=6000/(6000/1000+4)+300
▭(b_f=900 mm)

S8. Ans.(b)
Sol.

Distance between signal post and fort – A (h_1 )
h_1=0.06735 d_1^2
3000=0.06735 d_1^2
▭(d_1=211.05 km)
Distance between signal post and fort – B (h_2 )
h_2=0.06735 d_2^2
5000=0.06735 d_2^2
▭(d_2=272.46 km)
Hence, distance between both forts (d) = d_1+d_2
d=211.05+272.46
▭(d=483.51 km≃480 )

S9. Ans.(a)
Sol. Given, bearing stress (f_b )=300 N\/mm^2
Thickness of plates (t)=12mm
Nominal dia of rivet (ϕ)=22mm
Gross dia. of rivet (d)=ϕ+1.5
=22+1.5
=23.5 mm.
Bearing strength of rivet (P_b )=d×t×f_b
=23.5×12×300
=84600 N
▭(P_b=84.6 KN.)

S10. Ans.(d)
Sol. Given, ϵ_x=+800×10^(-6) cm\/cm
ϵ_y=+400×10^(-6) cm\/cm
ϵ_z=-1200×10^(-6) cm\/cm
Volumetric strain =∆V/V=ϵ_x+ϵ_y+ϵ_z
=(800×10^(-6) )+(400×10^(-6) )+(-1200×10^(-6) )
=0

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